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A193910 Leap centuries in the revised Julian calendar. 6
2, 6, 11, 15, 20, 24, 29, 33, 38, 42, 47, 51, 56, 60, 65, 69, 74, 78, 83, 87, 92, 96, 101, 105, 110, 114, 119, 123, 128, 132, 137, 141, 146, 150, 155, 159, 164, 168, 173, 177, 182, 186, 191, 195, 200, 204, 209, 213, 218, 222, 227, 231, 236, 240, 245, 249 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Terms divided by 100, e.g., 29 indicates year 2900, which is a revised Julian leap year, but not a Gregorian leap year. Values below 20 are "proleptic" (only based on the formula).
LINKS
Claus Tøndering, Frequently Asked Questions about Calendars, Don't the Greeks do it differently?
FORMULA
a(n) = a(n-2) + 9. - Charles R Greathouse IV, Aug 09 2011
a(n) = 2 or 6 (mod 9).
For all positive integers n, a(n) = (1/4)*(18*n-17*(-1)^n-11), which implies a(2*n-1) = 9*n-3 and a(2*n) = 9*n-7. - Farideh Firoozbakht, Oct 08 2014
G.f.: x*(2 + 4*x + 3*x^2)/((1 + x)*(1 - x)^2). - Philippe Deléham, Nov 30 2016
EXAMPLE
20 mod 9 is 2; 2000 was a leap year in the revised Julian calendar.
24 mod 9 is 6; 2400 and 2000 also happen to be Gregorian leap years.
28 is the first integer greater than 16 only contained in A008586.
29 is the first integer greater than 16 not contained in A008586.
MATHEMATICA
Table[1/4 (18 m - (-1)^m - 11), {m, 56}] (* Farideh Firoozbakht, Oct 08 2014 *)
PROG
(Rexx) do C = 0 to 250; J = C // 9; if J = 2 | J = 6 then say C; end C
(PARI) a(n)=(9*n-5)\2 \\ Charles R Greathouse IV, Aug 23 2011
CROSSREFS
A008586 enumerates "Gregorian leap centuries" (N mod 4 = 0).
A193879 enumerates all differences from A008586.
Cf. A274406: numbers congruent to {0, 8} mod 9; A301451: numbers congruent to {1, 7} mod 9. This sequence lists the numbers congruent to {2, 6} mod 9.
Sequence in context: A299637 A190889 A020966 * A215918 A078476 A217687
KEYWORD
nonn,easy
AUTHOR
Frank Ellermann, Aug 09 2011
STATUS
approved

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Last modified April 25 03:15 EDT 2024. Contains 371964 sequences. (Running on oeis4.)